In physics we learn that accleration is a vector quantity parallel to the radius and orthogonal to the velocity. With the embedding $\mathbb{S}^1 \hookrightarrow \mathbb{R}^2$ and the induced riemannian metric, this means that the acceleration vector at a point $\mathbf{p} \in \mathbb{S}^1$ is in the fiber of the normal bundle at $\mathbf{p}$. In particular, the line through $\mathbf{p}$ and the origin.
However, technically, acceleration is in the fiber of the second tangent bundle $T_{(p,v)}T\mathbb{S}^1$, $v$ the velocity at $\mathbf{p}$. I must be missing something because this space doesn't appear to contain the normal fiber above.
Can someone point out what I'm misunderstanding?
Suppose $ M $ is a Riemannian manifold, and $ \gamma: \mathbb{R} \rightarrow M $ a path in $ M $. Recall that the acceleration of the path $ \gamma $ is a map $ a: \mathbb{R} \rightarrow TM $ defined as follows: the acceleration $ a(t) $ at time $ t $ is given by extending $ \dot{\gamma} $ to a vector field $ \dot{\gamma}' $ and then restrict $ \nabla_{\dot{\gamma}} \dot{\gamma'} $ to $ T_{\gamma(t)}M $. So, I really mean to say is that the acceleration is also in $ TM $, not $ T_{(p,v)} TS^1 $.
If you want to think of $ \gamma $ as an immersion in $ M $, then $ T_{\gamma(t)}M = T_{\gamma(t)}\gamma \oplus N_{\gamma(t)} \gamma $ where $ N\gamma $ is the normal bundle. And in particular, if $ \gamma: \mathbb{R} \rightarrow \gamma $ is a geodesic in $ \gamma $ (which is what you mean by uniform motion), then the acceleration of $ \gamma(t) $ will be in the normal bundle.